DARPA, the research arm of the Department of Defense in the US, staged the Red Balloon Challenge in 2009. Competitors had to locate ten red weather balloons that had been tethered at random locations across the US.

The intention was not that one person drive around the country with a pair of binoculars. Rather, I might ask all my friends on Facebook to look out for a red balloon and tell me if they saw one. They might then ask all their friends, and so on.

The winning team from MIT found all of the balloons in just nine hours using this type of strategy. But to encourage participation they offered $2,000 to the first person to send them the co-ordinates of a balloon. On its own this may not have been very efficient. So, crucially, they also offered $1,000 to whomever recruited that person to the challenge, and $500 to whomever recruited that person to the challenge, and so on.

One interesting question mathematically is how much money did the MIT team stand to lose? The Red Balloon Challenge offered a prize of $40,000.

In principle, the sender of a winning set of co-ordinates might have been been at the top of a long line of recruiters. Doesn’t this mean that the MIT team risked an enormous payout?

Well, no. Consider a recruitment chain of seven people. The total payout would be:

$2,000+$1,000+$500+$250+$125+$62.50+$31.25 = $3,968.75

The seventh payment of $31.25 is pretty small. If there were more people in the chain, their payments would be even smaller. Nonetheless, lots of small amounts can quickly add up to a large amount.

Suppose there were 17 people in the chain. The total payout would be $3,999.97 (to the nearest cent). The seventeenth person would have got 3¢.

Even so, there could have been many more people in the chain, and might not the total have slowly grown to an unaffordable amount?

Suppose the total amount payable is T and that there are infinitely many people in the chain. Then,

T = 2000 + 1000 + 500 + 250 + ...

Now multiply both sides of this by 2:

2T = 4000 + 2000 + 1000 + 500 + ...

Finally subtract the first equation from the second:

2T – T = 4000 + 2000 + 1000 + 500 + ... – (2000 + 1000 + 500 + ...)

The terms in the brackets cancel out all of the terms at the beginning except the first one. Thus we end up with T = 4000.

So even if there had been infinitely many people in the chain, the total payout would have been $4,000. Since there were ten balloons that gives a grand total of $40,000 which was the value of the prize. MIT were certain to make at least a little profit, provided they actually won the prize. Since the kudos of winning was worth more than the prize, their investment was well worth the risk.

Series such as this one are called geometric series. One of the earliest examples of them was Zeno’s dichotomy paradox. For me to walk 4000 metres, I first have to walk 2000m, i.e. half the total distance. I then have to walk 1000m, i.e. half the remaining distance. Then 500m. Then 250m. And so on. No matter how far I've travelled there's always half the remaining distance left to go. So I have infinitely many stages to complete and will therefore never get to the end of them.

The flaw in the argument is that last sentence. It assumes that the infinitely many stages of the journey will take infinitely long to get through. But they won't.

Suppose I walk at 1m/s. Then the first stage will take me 2000s. The second will take 1000s. The third 500s, and so on. So the total time taken will be T = 2000 + 1000 + 500 + ... We now know that this adds up to 4000. So I can finish the journey in a finite amount of time. (Indeed the 4000 seconds you would expect me to need to walk 4000m at 1m/s.)